Question

The value of $\frac{(-1{)}^{13}}{2^3}+\frac{2^3-1^{10}+3^2}{3^2-2^2}+{\left(\frac{7}{11}\right)}^3÷\frac{98}{121}$ is:

• A. $3\frac{173}{440}$
• B. $3\frac{17}{44}$
• C. $3\frac{137}{440}$
• D. None of these

Answer 1

The correct option is A $3\frac{173}{440}$

$\frac{(-1{)}^{13}}{2^3}+\frac{2^3-1^{10}+3^2}{3^2-2^2}+{\left(\frac{7}{11}\right)}^3÷\frac{98}{121}$

$=\frac{-1}{2\times 2\times 2}+\frac{2\times 2\times 2-1+3\times 3}{3\times 3-2\times 2}$ $+\frac{7\times 7\times 7}{11\times 11\times 11}\times \frac{121}{98}$

$=\frac{-1}{8}+\frac{8-1+9}{9-4}+\frac{7}{22}$

$=\frac{-1}{8}+\frac{16}{6}+\frac{7}{22}$

$=\frac{-55+1408+140}{440}$

$=\frac{1493}{440}$

$=3\frac{173}{440}$